Let $M\subset\mathbb{R}^n$ be measurable, $f\colon M\to\mathbb{R}$ continuous, bounded. Claim: $f$ is Lebesgue-integrable.
I was able to prove it for $M$ additionally bounded. How to reduce the arbitrary case to $M$ bounded?
edit: With $M\subset\mathbb{R}^n$ measurable we mean that the characteristic function of $M$ is Lebesgue integrable
My attempt: Since $f$ is continuous and bounded, $f(A)$ is bounded, so that $f(A)\subset [a,b]$. Now I don't know how to chose a suitable partition of $[a,b]$ to define step functions such that I can apply Beppo Levi's monotone convergence theorem for Lebesgue integral.
If $U$ is open in $\mathbb R$ then $f^{-1}(U)$ is open $M$ which means there is an open set $V$ in $\mathbb R^{n}$ such that $f^{-1}(U)=M\cap V$. This makes $f^{-1}(U)$ a measurable set. Hence $f$ is measurable. Integrability is not true in general. If $M=\mathbb R$ and $f(x)=\frac 1 {1+|x|}$ then $f$ is not integrable.
If $M$ has finite Lebesgue measure and $f$ is bounded by $C$ then $\int|f| \leq C\lambda (M) <\infty$ where $\lambda$ is the Lebesgue measure, so $f$ is Lebesgue integrable.